Question

can anyone prove the HERON'S FORMULA...
who will prove,will be a brainlist

Answer 1

In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria,[1] gives the area of a triangle when the length of all three sides are known. Unlike other formulas, there is no need to calculate other distances in the triangle first.


Heron's Formula -- An algebraic proof

The demonstration and proof of Heron's formula can be done from elementary consideration of geometry and algebra. I will assume the Pythagorean theorem and the area formula for a triangle


where b is the length of a base and h is the height to that base.



We have


so, for future reference, 

2s = a + b + c 
2(s - a) = - a + b + c 
2(s - b) = a - b + c 
2(s - c) = a + b - c

There is at least one side of our triangle for which the altitude lies "inside" the triangle. For convenience make that the side of length c. It will not make any difference, just simpler.


Let p + q = c as indicated. Then 

Answer 2

In a triangle, the altitude to a side is equal to the product of the sine of the angle subtending the altitude and aside from the angle to the vertex of the triangle.

Heron's Formula

For Triangle in the above picture, the altitude to side c is   b sin A or  a sin B

Here the area of the triangle will be

Area= ½(bc sinA)

Now we look for a substitution for sin A in terms of a, b, and c. It is readily (if messy) available from the Law of Cosines

cos A= (b2+c2-a2)/2bc

and substitution in the identity

sin A= √1-(cos A)2

sin A= √1-{(b2+c2-a2)/2bc}2

Factor (easier than multiplying it out) to get

sin A= √{1- (b2+c2-a2)/2bc} {1+ (b2+c2-a2)/2bc)}

On solving the variables we get,

sin A= √[{2bc-(b2+c2-a2)}/2bc] [{2bc + (b2+c2-a2)}/2bc}

sin A= (1/2bc) √{2bc-(b2+c2-a2)} {2bc + (b2+c2-a2)}

sin A= (1/2bc) √{a2– (b2-2bc+c2)} {(b2+2bc+c2)-a2}

Factor

sin A= (1/2bc) √{a2– (b-c)2} {(b+c)2-a2}

Factor and rearrange

sin A= (1/2bc) √(a+b+c) (-a+b+c) (a-b+c) (a+b-c)

Now where the semi-perimeter s is defined by

s= (a+b+c)/2

the four expressions under the radical are 2s, 2(s – a), 2(s – b), and 2(s – c).   So,

sin A= 2/bc √s (s-a) (s-b) (s-c)

Since,

Area = ½(bc sin A)

we have.

A= √s (s-a) (s-b) (s-c)

The Algebraic Proof

Let us consider the below triangle given in the image

Let a, b, c be the lengths of the sides of the above triangle and h be the height to the side of the lenghth c.

We know,

Semi-perimeter of a triangle (s) = (a+b+c)/2

So,

2s = a + b + c

2(s – a) = – a + b + c

2(s – b) = a – b + c

2(s – c) = a + b – c

There is at least one side of our triangle for which the altitude lies “inside” the triangle.             For convenience make that the side of length c.

Heron's Formula

We need to express h in terms of a, b, c and then substitute h in the formula of area of the triangle, which is A= 1/2 (ch)

Let p + q = c as indicated. Then,

h2 + p2= a2

and

h2 + q2 = b2

since,

q= c – p

so, q2 = (c – p)2

and

q2 = c2 – 2cp + p2

Adding h2 to each side, we get

h2 + q2 = h2 + c2 – 2cp + p2

Substituting the values, we get

b2 = a2 – 2cp + c2

To solve p, we get

p = (a2 + c2 – b2) 2c

 

Now as we know, h2 = a2 – p2 substituting the values for p we get the entire expression in terms of a, b, c.